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Tugas 3
- 1.
TUGAS 3
NAMA KELOMPOK: 1.SUSANDI 2.RAFIS 3.GERIAN
KELAS : 1EA
LATIHAN 7.1
1. β« 100 ππ₯ = 100π₯ + π
2. β« 6π₯ ππ₯ = 3π₯2
+ π
3. β«( 3π₯2
+ 4π₯ β 5) ππ₯ = π₯3
+ 2π₯ β
5π₯ + π
4. β«( π₯2
+ 1)β π₯ ππ₯ =
2
7
π₯2
1
+
2
3
π₯2
3
+ π
5. β«( π₯ π
+ π π₯ ) ππ₯ =
π₯ π+1
π+1
+ π π₯
+ π
6. β«(10π₯ + 30)3
10 ππ₯ =
(10π₯+30)4
4
+ π
7. β«(π₯2
β 3)4
2π₯ ππ₯ =
(π₯2β3)5
5
+ π
8. β«(π ππ2
π₯cos π₯) ππ₯ =
π ππ3
3
π₯ + π
9. β« π₯2
β π ππ π₯3
ππ₯ =
β cos π₯3
3
+ π
10. β« πΌπ π₯ ππ₯ = π₯ πΌπ π₯ = π₯ πΌπ π₯ β π₯ +
π
penyelesaian :
1. =
π
ππ₯
(100π₯ + π) = 100π₯ + π
2. =
π
ππ₯
(3π₯2
+ π) = 6π₯ + 0 = 100
3. =
π
ππ₯
(π₯3
+ 2π₯2
β 5π₯ + π) = 3π₯2
+ 4π₯ β 5 + 0 = 3π₯2
+4x-5
4. =
π
ππ₯
(
2π₯
1
2
7
+
2π₯
3
2
3
+ c)=
2π₯
β1
2
14
+
6π₯
1
2
6
+ 0 =
1π₯
β1
2
7
+ π₯
1
2
5. =
π
ππ₯
(
π₯ π+1
π+1
+ π π₯
+ π) =
π+1.π₯ π+1β1
1π1β1+0
+ π₯π π₯β1
+ π =
π+π₯ π
1
+ π₯π π₯β1
+ 0 = π + π₯ π
+
π₯π π₯β1
6. =
π
ππ₯
(10π₯+30)4
4
+ π =
10π₯4
4
+
304
4
+ π =
40π₯3
4
+ 0 = 10π₯3
7. =
π
ππ₯
(π₯2β3)5
5
+ c =
π₯10
5
β
35
5
+ π =
10π₯9
5
+ 0 = 2π₯9
8. =
π
ππ₯
(
π ππ3
3
π₯ + π) =
π ππ2 π₯
3
.
sin π₯
3
+ 0 =
(1βπππ 2 π₯) .
3
sin π₯
3
9. =
π
ππ₯
(
β cos π₯3
3
+ π) =
β3 sin π₯2
3
+ 0 = βsin π₯2
10. =
π
ππ₯
( π₯ πΌπ π₯ β π₯ + π) = π₯.
ππ₯
π₯
β 1 =
π₯β1ππ₯
π₯
- 2.
LATIHAN 7.2
1. β«8ππ₯
2. β«
3
4
ππ₯
3. β«9.75 ππ₯
4. β«β3ππ₯
5. β«(
β40
3
β10+15
)ππ₯
6. β«16 β2 ππ‘
7. β« π2
ππ₯
8. β«2π ππ
9. β«β21ππ’
10.β«
6
π
ππ₯
Penyelesaian :
1. = 8x+c
2. =
3
4
π₯ + π
3. = 9π₯. 75π₯ + π
4. = β3 x+c
5. =
40π₯
2
3
10π₯
1
2+15π₯
+ π
6. = 16π‘. β2 π‘ + c
7. = ππ₯2
+ c
8. = 2π. ππ + c
9. = -21 u + c
10.=
6π₯
ππ₯
+ c
- 3.
LATIHAN 7.3
1. β«π₯5
ππ₯
2. β« β π₯34
ππ₯
3. β« π₯β2
ππ₯
4. β«
1
π₯2
ππ₯
5. β« π‘100
ππ‘
6. β« π’2π
ππ’
7. β«
1
βπ₯
ππ₯
8. β«
π₯5
π₯2
ππ₯
9. β« πβ1
ππ
10.β«
1
π‘
ππ‘
Penyelesaian :
1. =
π₯6
6
+ π
2. = β« π₯
3
4 ππ₯ =
π₯
7
4
7
4
+ c =
4π₯
7
4
7
+ π
3. =
π₯β2+1
β2+1
+ π
4. =β« π₯β2
ππ₯ =
π₯β1
β1
+ π = β
1
π₯
+ π
5. =
π‘101
101
+ π
6. =
42π+1
2π+1
+ π
7. = β« π₯
β1
2 ππ₯ =
π₯
1
2
1
2
+ π =
2π₯
1
2
1
+ π
8. =
π₯6
π₯3
+ π
9. =
πβ1+1
β1+1
+ π = β
- 4.
10.β« π‘β1
ππ‘ =
π‘β1+1
β1+1
+π = β
LATIHAN 7.4
1. β« π π‘
ππ‘
2. β« π20π₯
ππ₯
3. β« π ππ₯
ππ₯
4. β« π0,25π₯
ππ₯
5. β« π
π₯
5 ππ₯
6. β« πβ3π₯
ππ₯
7. β«4 π₯
ππ₯
8. β«23π₯
ππ₯
9. β«1000,25π₯
ππ₯
10.β« π
π₯
5 ππ₯
Penyelesaian :
1. = π π‘
+ π
2. =
1
20
π20π₯
+ π =
π20π₯
5
+ π
3. =
1
π
π ππ₯
+ π =
π ππ₯
π
+ π
4. =
1
0,25
π0,25π₯
+ π =
π0,25π₯
0,25
+ π
5. =
1
π₯
5
π
π₯
5 + π =
5
π₯
π
π₯
5 + π
6. =
1
β3
πβ3π₯
+ π =
πβ3π₯
β3
+ c
7. =
1
πΌπ 4
4 π₯
+ π =
4 π₯
πΌπ 4
+ π
8. =
1
3 πΌπ 2
23π₯
+ π =
23π₯
3 πΌπ 2
+ π
9. =
1
0,25 πΌπ 100
1000,25π₯
+ π =
1000,25π₯
0,25 πΌπ 100
+ π
10.=
1
π₯
5
π
π₯
5 + π =
5
π₯
π
π₯
5 + π
- 5.
LATIHAN 7.5
1. β«cosπ£ ππ£
2. β«sin(
1
2
ππ₯)ππ₯
3. β«cos(18) ππ₯
4. β« π ππ2
(β3π₯)ππ₯
5. β« ππ π2
(2,5) ππ₯
6. β«sec (
5
6
π₯)tan(
5
6
π₯)ππ₯
7. β«csc
x
3
πππ‘
π₯
3
ππ₯
8. β«csc(ex) cot(ππ₯)ππ₯
9. β«sin 3π ππ
10.β«cos(25ππ₯) ππ₯
Penyelesaian :
1. = Sin v + c
2. = β
1
1
2π
cos (
1
2
ππ₯) + π = β2πcos (
1
2
ππ₯) + π
3. =
1
8
sin(18π₯) + π
4. =
1
β3
tan(β3π₯) + π =
tan(β3π₯)
β3
+ π
5. = β
1
2,5
cot(2,5 π₯) + π =
βcot(2,5π₯)
2,5
+ π
6. =
1
5
6
sec(
5
6
π₯) + π =
6
5
sec (
5
6
π₯) + π
7. =
1
1
3
csc(
x
3
) + c = 3 csc (
π₯
3
) + π
8. = β
1
π
csc (ex) + c =
β csc( ππ₯)
π
+ π
9. = β
1
3
cos(3π) + π =
βcos(3π)
3
+ c
10.=
1
25π
sin(25ππ₯) + π =
sin(25 ππ₯)
25π
+ c
- 6.
LATIHAN 7.6
1. β«
1
1+π2
ππ
2.β«
ππ₯
β16βπ₯2
3. β«
1
49+π₯2
ππ₯
4. β«
ππ‘
0,25+π‘2
5. β«
ππ’
βπ’2(π’2β1)
6. β«
1
|π₯|βπ₯2β41
ππ₯
7. β«
1
β
81
100
βπ₯2
ππ₯
8. β«
1
π2+π₯2
ππ₯
9. β«
ππ‘
βπ‘2(π‘2β
1
4
)
10.β«
1
|π₯|βπ₯2β7
ππ₯
Penyelesaian :
1. = π‘ππβ1
π + π = βπππ‘β1
π + π
2. = β«
1
β42ββπ₯2
ππ₯ = π ππβ1
(
π₯
4
) + π
3. = β«
1
β492+π₯2
ππ₯ =
1
β49
π‘ππβ1
(
π₯
β49
) + π
4. = β«
1
β0,252+π‘2
ππ‘ =
1
β0,25
π‘ππβ1
(
π‘
β0,25
)+ π
5. = β«
1
|π’|βπ’2ββ12
ππ’ =
1
1
π ππβ1
(
π’
1
) + π = π ππβ1( π’) + π
6. = π ππβ1
(
π₯
41
) + c = - ππ πβ1
(
π₯
41
) + c
7. = β«
1
β
(9)2
10
βπ₯2
ππ₯ = π ππβ1
(
π₯
9
10
) + π =π ππβ1
(
10π₯
9
) + π = βπππ β1
(
10π₯
9
) + π
8.
1
π
π‘ππβ1 π₯
π
+ π = β
1
π
πππ‘β1
(
π₯
π
)+c
- 7.
9.= β«
1
| π‘|βπ‘2β
(1)
2
2
ππ₯ =
1
1
2
π ππβ1
(
π‘
1
2
) + π = 2π ππβ1(2π‘) + π
10.π ππβ1
(
π₯
7
) + π = βππ πβ1
(
π₯
7
) + π